integration - Calculate $int_{-infty}^{infty} e^{-x^2}cos (ax) dx$ using Taylor series cosine

Let $a > 0$. Im trying to show that $\int_{-\infty}^{\infty} e^{-x^2} \cos (ax) dx = \sqrt{\pi}e^{-\frac{1}{4}a^2}$. I'm taking a course on measure theory, and I want to prove this using the Lesbesgue's Dominated Convergence Theorem. The idea is to write
$$
\cos ax = \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} (ax)^{2m},
$$

so assuming we may switch the integral and summation using the above mentioned theorem (which I haven't proved yet), we get
\begin{align*}
\lim_{R\to\infty} \int_{-R}^{R} e^{-x^2} \cos (ax) dx
&= \lim_{R\to\infty} \int_{-R}^R e^{-x^2} \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} (ax)^{2m} dx\\
&= \lim_{R\to\infty} \sum_{m=0}^\infty \frac{(-1)^m\cdot a^{2m}}{(2m)!} \int_{-R}^R e^{-x^2} x^{2m}dx.
\end{align*}
But now we are left with
$$
\int_{-R}^R e^{-x^2} x^{2m} dx
$$

for $m\in \mathbb{N}$. I know that for $m=0$ this integral doesn't have a nice solution, and I doubt it has a nice solution for $m\geq 1$. Is it possible to solve this problem using this route, or is it a dead end?

I'm aware that there are other methods to solve this, see Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$.

Edit: I solved it using Lebesgue's Dominated Convergence Theorem (LDCT). First, I proved that the improper Riemann-integral was equal to the Lebesgue integral, and then I switched the summation and integration using LDCT. I ended up with $\int_{\mathbb{R}} e^{-x^2}\cdot x^n d\lambda^1$, which I could calculate using an improper Riemann-integral. I ended up with the Taylor series of $\sqrt{\pi}e^{-\tfrac{1}{4}a^2}$.